On The Harmonic Series

ON A CRUCIAL VIEW OVER THE HARMONIC SERIES This essay is a simple inference into the harmonic series and it contains a set of derived formulae expressing some of the relationships existing between the terms of this series under consideration, and their convergence towards one another, and in particular 1. If we consider the geometric series; 

1

i 1

2





 i 1

1



1

i 1

3



Now consider the series;

1 3

=

3

i

i

1

i



1

i 1

2



i

then clearly;

= 1.

then clearly;

1

1 3

31 . 2 2 3

=

Similarly : 1 1  4   i 1 i 1 4 1 4 

1 3

4 1 3 4

Continuing in this way, we observe that 

 i 1

1

k

i



1 k 1

k  2. ……………………..(0)

From the above observation , we can observe that ;



1

i 1

2



i

+



1

i 1

3



That is ;    1  j  =    i  2  j 1 i 

i



+

1

w w1



1

i 1

4



i

+…+



 i 1

k

  1 +…+… =    j  = i  2  j 1 i  

1 i

……………………1



1

w w1

=  (1) .

  1   j  =  (1) ……………………2    i  2  j 1 i  

 (1) Which is the harmonic series is divergent (as can easily be proved with the test of Oresme). From this short-hand we can now set forth an expression for the general “   function ”.

  1  j  =    i  2  j 1 i  

Since :

  1     j  i  2  j 1 i  

s



w1





=

1

w,

w1

1 s

w

=  (s) .

That is ;

s

  1   j  =  (s) ……………………3.    i  2  j 1 i  

From where ;     1 s         i j   =  s 1 i  2  j 1    k

k

 s

=  (1) +  (2) +…+  (k) =

s1

k

  ( z ) ……………4 z 1

ON A DECOMPOSITION OF THIS SERIES. Due to the fact that; 

 i 1

1

k

i



1 , k  2. k 1



1

i 1

2





i

=1 ,



1

i 1

3



i

=

1 . 2

Putting the above two together, we condense as; j

  1   i  = 1 ………………………..5  j 1  i 1 3  

Similarly, invoking the formula;





j

1 1 = , i2 . j i i 1

k

    1 j        =1   j 1  i1 4 i   k 1   

Then we can deduce that;

Continuing in this way, we can arrive at the beautiful formula below; xn 1 x2             1    ...      ...    x  n  1     xn 1  xn 1 1   x2 1  x1   

xn

= 1, with n  1 . …………………..6

Upon considering equation 6, we see that ;

lim n 

xn 1 x2               1  ...    ...      x     xn 1  xn 1 1 x2 1  x 1 n  1       

xn

xn 1 x2             1    ...      ...    x  lim    n  1 n  xn 1  xn 1 1 x2 1  x 1        

=

lim 1 = 1. That is ; n 

xn



Since each term of the harmonic series; S= 1 

= 1 …………………6(#)

1 1 1   ...   ... 2 3 k

Can in some way be represented as a geometric series of the term(s) that lag(s) it, then naturally, any term can be represented as a linear combination of the geometric series of any number of terms after it. As an example, we see that; 

1

i 1

2





1

i 1

3



= 1, i

i

=

1 , then; 2



1

i 1

3



i

+

1 =1. 2

Similarly, by equation O, we obtain that; 

1

i 1

3



i

+



1

i 1

4



i

+



1

i 1

7



i

= 1.

From where it is possible to find more terms of the harmonic expressed as geometric series such that their sum equals to 1. Since the harmonic series has infinitely many terms, there are infinitely many combinations possible such that the sum equals to 1. There arise the natural need to find convenient a notational shorthand for these iterated sums over infinity, which shall be the main way data shall most of the time be handled. Since k

    1 j   1   i  =      i   =…=   j 1  i1 4   j 1  i 1 3  k 1   



j



xn 1 x2             1    ...      ...    x     n  1 xn 1  xn 1 1 x2 1  x 1        

xn



We see that a function “  ” can be introduced as;  :   i1

i2          in    :    ()     ...   ...  ………………………7      i1  i2  in    

The latter equation is good looking true but very inconclusive, as “n” is arbitrary. Let a function we decide to name the “sum iterate” be defined as follows:



k i

(  ) where “k” is the cardinality of the fold of the sum of the value “  ”.    .

= 1.

I now define the “sum iterate”.



k i

:      : , k   

if



i

,k=0

if

,k=1

i

in 1 i2            1         ...    i  ...    i2 1  i 1 in 1  in 1 1        

in

if , k  n

i 1 i2           1   With  i        ...    i  ...      i2 1  i1      i 1  i 1 1      the sum-iterator. 

i

being the “  ” fold sum of, “  ” with

Let the convention  i ()   i () be adopted. With, the sum-iterator notation, all equations from 1. to 6. can be rewritten as; 1





  i ( 1i )    i ( 1i )..............1 i 2 i 2 1

s

  1    i      j  

  ( s ) ………. 3

j 2

1   which is different from i s  j 2 j  establish the following beauty;

3 . Can be rewritten as;





s

  1    i      j  

  (s) =

j 2

p

And thus,



 j 2



 j 2

i

1  s  …………………. 3 j 

 1  p      s  ………..4 i s     s 1

     j s 1





 j 2

s i

1   . From where we  j

And hence;



i



i

2

m

1   = 1 …………. 5  3  1     1 ………… 6  m 1

m  1.

Observe that, for there to be a chance that  i (  ) be finite, the parameter “  ” of the sum iterate must be such that; 0 <│  │< 1. Thus 6(#) can be written as; k

lim  k 

k i

( )  1



1 . k 1

**THE FOLD OF CARDINALITY IS THE ORDER OF THE POWER TO WHICH THE GEOMETRIC SERIES IS CONSIDERED** Lemma 1:

The sum iterate is convergent at the parameter “  ” with finite value if the parameter “  ” obeys the condition; │  │< 1 and fold cardinality is 1. Proof of lemma: By the definition of the sum iterate; in

in 1 i2            1     n  i ()      ...    i  ...   ………..1 i2 1  i 1 in 1  in 1 1        

i): Suppose   1

in

in 1 i2               n  if  =1 , then  i (1) =    ...   1x  ...      i2 1  i1     in 1  in 1 1      But since the innermost sum tends to infinity as it is the equivalent of summing infinitely many “ones”, and hence the infinite sum of infinite exponentials is naturally infinite. So 



n i

(1) =   . For all n  1.

Now, if  >1, and because  and hence  are well ordered, then  i ( p ) > i (1) for p>1. but, since  i (1) =   it implies   1.



i

( p ) =   . And as such,  i () =   for

ii): Suppose  < 1.   i () =

 . i



But if use is made of the geometric series structure S n =  ar n1 n

 with “a” being the first term, then we obtain; i

S n = a+ ar + ar 2 +…+ ar k +… Now r S n = ar + ar 2 +…+ ar k 1 +… Due to the fact that: 0<  < 1. Then, r<1 and hence r S n < S n Naturally, (S n - r S n ) > 0. And as such S n (1- r) = a (1- r n ) and hence

a (1  r n ) for all n > 0. As r < 1. And hence, r n << 1. (1  r ) Taking limits as n   then;

Sn =

lim r

n

= 0.

n 

From which fact we conclude that; Hence



i

0<│  │<1. And hence,

() is convergent to



i

() =

 .<   . 0<│  │< 1. 1 

 and id finite since │  │< 1.and as such , 1 

 < . 1 

Q.E.D The above lemma merely proves the finiteness of the value of  i () only for 0<│  │<1,  > 0.and thus says nothing about the behaviour of the sum iterate over cardinalities of folds in relation to the parameter it takes. The study of the sum iterate in this essay shall be focused as said earlier on the terms of the harmonic series that are less than 1.

If we go back to equation 6. above, then we observe that



m i

 1    =1 and thus  m 1 

1  m 1  naturally  i   =   since it is the equivalent of summing infinitely many ones.  m 1 We can set forth the following corollary;

COROLLARY 1 If the parameter of the sum iterate is such that; 0 <  < 1, then the following are true; i)



ii)



card ( 1 ) i

() =   .

card ( 1  k ) i

() =

1

; k= 1

a , 0 < a <1; k>1 and card( 

1

- k) < card( 

1

- 1)   s.t 

1

 1.

PROOF OF COROLLARY: i) 1   1 1 1 Let S  (1) = 1, , , ,..., ,... chose  s.t   S  (1) . k   2 3 4

By hypothesis of the corollary, 0 <  < 1, and thus, we can define a subset S (1) of S  (1) s.t S (1)  S  (1) . And S (1) = S  (1) \{1}. Then for   S (1) , we have that;



card ( 1 ) i

() =

 (...(  

i

i1

i

card (  1 )

For   S (1) , we have 

1

card (  1 )

)...)i1 .



 2,and as such card( 

Hence, for the elementary case where card( 

1

1

)  card(2)=2. i.e card( 

1

)  2.

) = 2 we have ;

 = ½. 1 So that  i   = 2 2

Since    S (1)

j

   1 = 1j =   .      i 2 j 1  i 1 j 1  1 with   2 and hence   1/2, we see that card(  

card ( 1 )

and consequently ;  i

() =    card( 

1

)  2. End.

1

) is well ordered,

ii) a)



r.t.p

card ( 1 1) i

   S (1) , we have card( 

() = 1. 1

)  2 and hence card( 

1

-1)  1.

Elementary demonstration; For   S (1) , card( 

1

)  2.

Chose  = ½ with card(  But 

1

1 card (   1) 2 i

1

) = 2. Then, card( 

card (1 )  1  1 1  = i  = i   = 2 2 2

Chose  =1/3  S (1) , then card(  

1

1

1 card (   1) 3 i

1

-1) =1. 

1

i 1

2



i

= 1.

) = 3. and naturally card( 

card ( 2 )  1  1   =   = i 3 3

1

-1) =2.

j

   1 1   i  =  j =1.  j 1  i 1 3  j 1 2



Recursively, it can be shown inductively that this is consistent with equation 6. with the unequivocal fact;

 b)

r.t.p



card ( 1  k ) i

card ( 1 1) i

() =1. End

() = a ;

Let   S  (1)  naturally, card( 

1

0< a <1 , k>1

)2 1

1 Still, if the elementary case of  =1/2 is considered, with k=2 ,  (   - 2) =0. 2 0 1  Thus  i   = ½ (by the definition of the sum iterate), and 0 < ½ < 1 showing the 2 validity of the elementary statement. Now, let   S (1) = 1/3.  card(  1 ) = 3.But ,since card(  1 - k) < card(  1 - 1).

In which,case k>1. Let k = 2



1  i  3  =



1

i 1

3



1

=

i

1 and 0 < ½ <1. 2

If   S (1) =1/4.  card (  logical truth. k = 2 ;  i

( 4 2 )

K=3;



) = 4 and hence k= 2,3 will do the job by recurrence of

2 1  1   = i   = 4 4

( 4 3 ) i

1

j

  1  i  =  j 1  i 1 4 

1 1   = i  =  4 4





1

i 1

4



i



1

j 1

3



j

=

1 …………A 2

= 1/3 ………….B

Observing A. and B. we see that 0< ½, 1/3 <1 .Which is still in agreement with the statement. Hence, we have that for k = 2,3,4,… card(  1 -1 ) satisfies the assertion. Thereby, permitting us to close this proof by demonstration at this point. End. Q.E.D Haven established a few proofs , we can now proceed with the algebra of the sum iterator but not before setting forth the following theorem; THEOREM 1 If  i () is a sum iterate that takes parameter  ,for arbitrary (k-10 fold of cardinality, then, to move from the (k-1)st to the kth order of cardinality, it suffices to ( k 1) raise the result  i () to some power governed by some indice and then taking the infinite sum of the resultant geometric series. i.e without loss of generality;



k i

  

() =

k 1

( k 1) i



k

()

is mathematical shorthand of the theorem.

PROOF OF THEOREM By definition  i () = k

And thus  i () = p

 (...(   )...) ik



i1

ip

i1



i p

Assuming p > k. p, k   .

…………………..1

ik 

 (...(   )...) 

i1



i1

……………………2

Since 1    if we set p = k+1, then as k+1 > k is valid due to the fact that  is well ordered (or better still is an inductive set). Because p has been set to be equal to k+1, substituting in equation 2, we obtain;



( k 1) i

() =

 (...(  

ik 1

i1

ik 1

)...)i1 . Yet  i

( k 1)

() =



ik 1

(...(   1 )...)ik 1 i

i1

Since the subscript of the parameter of summation is irrelevant, we can as well do the change i k 1 to some “a” and obtain; ik i2           ( k 1)  ...  1  ...    (  ) =    i    i i2 1  i 1   ik 1 1 ik 1     

ik 1



………………3.

ik 1 i2             1  k But, by definition,  i () =     ...    i  ...     i2 1  i1      ik 1  ik 1 1     

ik



i

( k 1)

() =

 

ik 1

k i

()



k 1

 

=

a

() i k



a

as required. Q.E.D

At the of the proof of theorem 1, I think it is now safe, with the idea of the availability of the above theorem, corollary an lemma to attempt a journey in areas which, shall as I hope enhance understanding of the behaviour of the terms that make up the harmonic series in relation to the sum iterate concept and convergence to the value 1,thus bringing forth the beauty of the truths that emanate from some simple relationships existing between it’s terms. SUM ITERATE ALGEBRA Let  be the parameter of the sum iterate of arbitrary fold of cardinality. The aim here is to study the behaviour of the results when the parameter  for some fixed fold of cardinality varies, and also, when, the situation when the parameter  is fixed whiles the fold of cardinality is made to vary. Let some axioms be laid here; Since S (1)  S  (1) ,  let;



i

: S (1)  S (1) :





i

() For arbitrary fold of cardinality.

Impose the condition that;  Then;



p i

() , (p = 

1

- 1); 

1

 k  1………….*

1 1 = a 2 ,…, = a k 2 k This will ease the work to be done. We can now write the set as follows;

Let’s ordinate the set S (1) as follows: 1 = a 1 ,

S (1) = a1 , a2 , a3 ,..., ak ,... . We shall do the study on a case by case basis. VARYING THE PARAMETER WITH FIXED FOLD OF CARDINALITY Let the cardinality be set to zero i) Set k = 0. By the definition of the sum iterate; when k= 0, then  p  S (1) , we have that ;  i ( p ) = p. To enable insight into S (1) to be “smoother”, we shall work with the ordinated set S (1) . k

  a j  S (1) we have that  i ( a j ) = a j ,  j   But because k = 0, we have that; k



k i

( a j ) = a j ……………….8

 a natural consequence is that;

 a  = 1+ ½ + 1/3 + …+ 1/k + … =  q 1

k

i

j

j

 a  k

i.e



j

i

=  (1)

q

j

=  (1) k= 0 …………9.

Observe that, for k= 0,    S (1) ,  i () <   ,as   S (1) is s.t   1 and hence by the definition of the sum iterate at fold cardinality value zero, equation 8 is satisfied. k

If   S (1) ,    ,  i () =  and hence ,if   S ( S is some arbitrary subset of  , then ,if we ordinate the set S we obtain ; k

   =  k

j

i

j

1

+

2

+ …+ 

As an example, if S =  ,  

k

k

+… = k  k   .

   , with   

n(n  1) . j 2 j Quite simple and beautiful. And, this rule demonstrated above is valid, by the truths from which it was derived.

And hence



k

i

and  = 1,2,3,..., n, equals

ii)Set k= 1. By corollary 1, we have that  i () =   if   1. But our domain of study as k

established in the beginning of this phase of study, we imposed that   S (1) . Hence, clearly  i ( a1 ) =   ,  i ( a2 ) = a1 ,  i ( a3 )  a2 ,… if we continue in this way, then inductively it can be deduced that; k

k



k i

k

 j  2.

( a j )  a j 1

i.e



k i

( a j )  a j 1 k=1 j   …………………10.

Consequently, we have that, taking values in the ordinated set S (1) , we obtain;



k i

(a2 ) +  i (a3 ) +…+  i ( aq ) +…= k

k



1

x

=  (1) ………………11

x 1

iii)Set k=2 Still by corollary 1, we have that    S (1) ;



(a3 ) = a 1 =1,  i (a4 ) = a 2 ,…, and thus  i (a j ) = a j 2  j  3…………12 And as such;   1 k ( )    (1) ……………….13 a   j  i j 3 m 1 m A very simple and beautiful rule is establishing itself here. Observe that when k=0,we exclude no value in  ,and a sum of the subscript of the parameter of the sum iterate over the cardinality of  equates naturally the result to  (1) . Setting k=1, we sum the parameter of the sum iterate, over all subscripts that do not have the value 1 so as to be able to equate result to:  (1) . k

k

k

i

Continuing in this way, we observe that: For k=0

a j : j=1,2,3,…

i.e S (1)

K=1

a j : j=2,3,4,…

i.e S (1) \ a1

a j : j=3,4,5… i.e S (1) \ a1 , a2  . . . . . . a j : j=n,(n+1),(n+2),… i.e S (1) \ a1 , a2 ,..., an 

K=2 . . . K= n

And hence, as k tends to infinity, the set over which the validity of the parameter of the sum iterate is respected tends to the empty set  . From where we obtain the following truth; 



j  k 1

k i

(a j ) =  (1) ………………..13’

A mathematical shorthand for the fact just before equation 13’ is;

lim  k 

k i

( :    │   S

 (1)

  \  ak  ) =0………………14  k 

As for me, the most beautiful equation so far is equation 14. Since the infinite sum over the element of the empty set… (oops ! the empty set has no elements and thus the infinite sum over nothing) is zero, from where equation 14. And again equation 14 is a beautiful consequence of the constraints of validity of the sum iterate in the set S (1) . And as such; c      S  a : a S  ( 1 )  ( 1 )   p   =  ……………….15 p lim p   p    

II.VARYING THE FOLD OF CARDINALITY WITH FIXED PARAMETER Everything that is to be done here, will be done within the constraints of validity of the sum iterate w.r.t the set S (1) . From the definition of the sum iterate,



k i

() =  for k=0.     .

But, the domain under consideration here is S (1) and as such,    S (1) ,we have that;



k i

() =     S (1) . K=0.

For arbitrary   S (1) ,if we set k=1, then, a simple consequence is the fact that;

  i  ordinated S (1) …

** An aside; let the ordinated set S (1) be designated O( S (1) ) with the same properties as set above ** …choosing  1  O( S (1) ), then:



k i

(1 ) =  1 ; k=0 .   ; k  0.

Setting k  1, then , for  1  O( S (1) ) we have that;



k i

(1 ) =  1 ; k =0

 ; k 1

………………….A

Choosing  2  O( S (1) ), we have that :



k i

( 2 ) = 

1 ;k=1  ; k  2

Choosing 



;k=0

2

k i

 O( S (1) ), we have that :

3

(3 ) = 

…………………B

;k=0

3

 2 ;k=1 1 ;k=2  ; k  3

…………………….C

Observe that   j  O( S (1) ) ,generally, the following distribution is observed ;



k i

( j ) = 



j

; k=0

j 1

; k=1

 j 2 . . . 2 1 

; k=2 …………………….P ; k= j-2 ; k= j-1 ; k j

 j 1

Equation D is derived inductively on the ordinated set O( S (1) ) . 



A 

k 0 



B 

k 0

k i





C 

(1 ) =  1 + (   ) =   ……………….A(*)

k i

k 0

( 2 ) =  1 +  2 +(   ) =   …………….B(*)

k i

(3 ) =  3 +  2 +  1 +(   ) =   ……………….C(*)

 ( ) =  B(*)   ( ) =  C(*)   ( ) =  A(*) 

k

i k

1

1

;k=0

i k

2

2

,  1 ; k= 0,1

,  2 ,  1 ; k = 0,1,2 . . . . . . . . . . . . k P(*)   i ( p ) =  p ,…,  2 ,  1 ; k = 0,1,2,…,(p-1) 3

i

3

For arbitrary  p  O( S (1) ), the value of validity of the cardinality can only vary from o to (p-1). Hence it becomes necessary to define a set V which is the set of validity of the varying cardinality for an arbitrary   S (1) as; 

k

 O( S (1) ), we have the set of validity to be;

 h1  V=  k :  k  ……………..16  k 0 

But , for the elements to tend to  (1) ,the situation that the ordinality of the element chosen from O( S (1) ) tend to infinity so that the cardinality of the sum iterate too should tend to infinity and hence all the elements of the set S (1) will be generated thus bringing forth  (1) . Since

j 1

 k 0

k i

( j ) =  1 ,  1 ,…,  1 . For j  1………………..17

**The ordinality refers to the position ** It turns out that this is an infinite sequence of terms as j tends to   ,however, the aim here is to represent  (1) ,as a relationship derived naturally from the above essay. Hence we can step forth the following equation after considering equations 15,16,17.

 k 1 j    i (k ) = S (1) …………………..17, for k   and k= j+1 . lim k   j 0  And as such, we can deduce from equation that, because  i ,  j  O( S (1) ), with i

j

if i  j.  we have the equation;

 k 1 j   i (k ) =  (1) …………….18  lim k 1  j 0  ( is equivalent to saying that k goes from 1 to infinity) Before, let’s define the operator  as follows; Let D =

 D  with the D q

z

’s disjoint. Then;

q



:  D   :

 D  q

q



D

q

With D z  D.

q

  j Observing that  i (k )  k  ……….E  j 0 

  j  i (k )  k 1 ……………….F   j 1

  j  i (k )  k 2 …………..G   j 2

And hence We do generalize as;    j    p  =  (1)  ( )   With  i k   k q p i  j q  Because equation 17 is somewhat ambiguous, we can thus, with the help of the operator “  ” which generates the required series  (1) as;

 k 1 j   i (k ) =  (1) …………….18  lim k 1  j 0  Finally, let’s get the set of validity;

When k= 0, the set of validity is; V= S (1) \ S (1) \ 1 Setting k=1 , the set of validity is ; V = S (1) \ S (1) \ 1 ,2  Setting k=2, the set of validity is; V = S (1) \ S (1) \ 1 ,2 ,3  Generally, we can obtain the set of validity for fold of cardinality “z” The set of validity is; z

 z 1   z 1  V= S (1) \ S (1) \   j  =   j  .  j 1   j 1 

Considering the fact that if “A” is a set then A \A =  , but, if Bs.tB  A,  ; A \ B = A  B c Hence, A \A\B =A  ( A B c ) c = B. Lemma2: If “A” is a set (arbitrary), and “B” too is a set, then A\A\B =B. With A,and B, being real subsets. PROOF OF LEMMA

If A  B, then ;

A\A\B = A\(A\B) But, by De-Morgan’s rule, A \ B = A

 B implying that; = A  A  ( B )  =B. c

c c c A\(A\B) = A  ( A B c ) c Similarly, if we set B=A we have the same procedure to get to the same result, i.e A\A\B=A\A\A =A

Q.E.D And hence the set of validity required is;

lim

j1

 j 1 k   i ( j ) = S (1) ……………….19.  k 0 

ON THE TWO-DIMENSIONAL GEOMETRY OF THE ‘‘S  (1) ’’ SET

Define an axis (In the sense of the real “x-axis”).comprising only of the elements of “S  (1) ” and I wish to baptize it the “x- S  (1) ” axis. Define another axis orthogonal to the other (i.e in the sense of the real y-axis) which I wish to baptize the “y- S  (1) ” axis, still comprising only of the elements of the set “S  (1) ” . So that, we can analogously obtain a corresponding OXY plane. But in this case I wish to set forth the notation ; OXY S ( (1)) to indicate that the elements of the line are exclusively those from the set “ S (1) ”. Define areas bounded at the four corners by these points now of the form (x,y) (understood as point coordinates in the plane),as “regions” , or rather still, to put aside ambiguity, “vacuums” because positions are defined as combinations of the points and not of the areas. If we could, and can so, do draw parallel lines from the points on the axis, and then we shall notice we have a very beautiful picture and it is its properties I wish to investigate in this essay. Fist, we wish to extrapolate the “vacuum-values” relationships and from there attempt, fairly to deduce some possibly existing underlying rule governing the inter-relationships between the “vacuum-values”. Just by thinking about it then it becomes clear that plotting points on this plane will generate some symmetric picture, picture being symmetric about the “leading diagonal” of the areas, the areas about the points are described thus; ON THE FIRST COLUMN;  1  1  1 A(11) =    =  2  2  4 *1 The figures below the first column are such that the area of the “ith” figure is given by the relation; 1 1 1 1  1  1   1  . A(21) = (1- )(  ) =    =   = 2 2 3  2  6   12  4 * 3 Similarly; 1 1 1 1  1  1   1  A(31) = (1- )(  ) =    =   = . 2 3 4  2  12   24  4 * 6 Therefore I wish to conjecture that; 1 1 1 2  1  1  1  = * ) =   ……….20 A(i1) = (1- )(  2 i i 1  2  i (i  1)  4 i (i  1) ON THE SECOND COLUMN; On the second column, the interval of length is [1/2, 1/3], that is the range of the length is 1 1 1 ; (  )= . 2 3 6 Hence we have the following results; 1 1 1 1 1 A(12) = (1- )(  ) =   = 2 2 3  12  12 * 1

1  1    =  36  12 * 3 1  1    =  72  12 * 6 1 1  2   ………….21 I then conjecture that; A(i2) = * *  4 3  i (i  1)  Before I continue I wish to set forth an axiom; 1 1 1 1  )(  ) = 2 3 2 3 1 1 1 1 A(32) = (  )(  ) = 2 3 3 4

A(22) = (

AXIOM: The width of the (k-1)st rectangle equals the side of the kth square. Symbolically represented or, rather stated as;  (rk 1 )   ( S k ) ………..22 THE THIRD COLUMN; 1 1 1 1  1  1   1  A(13) = (  )(1- ) =    =   = . 3 4 2  2  12   24  24 *1 1 1 1 1 1  1  1   1  A(23) = (  )(  ) =    =   = . 3 4 2 3  6  12   72  24 * 3 1 1 1 1 1  1  1   1  A(33) = (  )(  ) =    =  . = 3 4 3 4  12  12   144  24 * 6 Generally; 1 2 1 1 2 * = * * ………….23 A(i3) = 24 i (i  1) 4 6 i (i  1) Remembering the sequence; ( an ) n = 1,3,6,10,15,…, n(n  1) /2,…, then we observe that in the set ; k (k  1) , with k   2 i 1 By virtue of the axiom above I can conjecture that:

O(( an ) n ) =

n

 a , with a i

k



UNDER THE “qth” COLUMN, THE “pth” AREA IS; A(pq) =

1 1 1 1 1 * * = * ………….24 4 a p aq 4 a p aq

with p, q   .

Now, since I want the “areas matrix”, to be somewhat consistent with “norms” I will advice that the “x-axis” be taken for the negative “x-axis”, and maintain the positive “y-axis” so that our resultant areas matrix be of the form of a coefficient matrix of a system of linear equations. Let A(pq) represent an entry in this area matrix , then I can rewrite this to fit norms as “ a pq ”.

By simple thought, we see that this area matrix is finite, yet, has infinitely many entries. Call the matrix “  ”  = ( a pq ). Claim; the matrix  is symmetric Proof;  = ( a pq ),   T = ( aqp ). It suffices to show that a pq = aqp . a pq =

1 1 1 1 * = * (since multiplication of real numbers is commutative). 4 a p aq 4 a q a p

= aqp .  ( a pq ) =  = ( aqp ) =  T .

Hence  =  T . Q.E.D We therefore obtain the area matrix as;  a11 a12 a13 ... a1n ... ...     a21 a22 a23 ... a2 n ... ....   ... ... ... ... ... ... ...     an1 an 2 an3 .... ann ... ...    ... ... ... ... ... ...   ...  am1 am 2 am3 ... amn ... ...    ... ... ... ... ... ...   ... It is true, you could wonder, how come the right most bracket is closed. But I’ll say the matrix, though by virtue of the set from which it was derived has an infinitude of entries, it is finite! Because;  e  S (1) , e  0,1 , and the diameter of this interval I = 0,1 < +  .and in fact;

 ( I )  1 <+  . So our matrix is square and has side of unit length. Hence, we have that;  Has finite dimension, that is, is finite in dimension (or length), and yet has infinitely many entries.  a11 a12 a13 ... a1n ... ...     a21 a22 a23 ... a2 n ... ....   ... ... ... ... ... ... ...     =  an1 an 2 an3 .... ann ... ...   ... ... ... ... ... ... ...    am1 am 2 am3 ... amn ... ...    ... ... ... ... ... ...   ...

A preview over the nature of the matrix  is given below; 1 1 1 1  1  ...   40 60  4 12 24  1 1 1 1 1  ...   12 36 72 120 180   1  1 1 1 1 ...  =   24 72 144 240 360   1 . . . . ...   40   . . . . . ...    ... ... ... ...   ... ... The most of the matrix can be written as chosen by simply making use of the AXIOM 1, the fact that the matrix is symmetric and by imposing as wished, restrictions to the dimensions of the matrix simply by choosing an interval;Y   in which the parameters p and q take values. That is, a sub-matrix can always be extracted from the matrix  by simply choosing s required to suit some purpose, a subset of the set of real numbers